# Air pressure problem

1. Apr 3, 2014

### edanzig

The question asks; "Assuming that the density of air is a constant 1.3 kg/m3 and that the air pressure is 1 bar, what is the pressure at the top of a 500 meter high skyscraper?"

I'm having trouble setting something up but here is what I do have;
Regarding liquids we know that F = (Rho)(height)(g) so I want to say that whatever "causes" the "air to push down" on the ground at 1 bar of pressure should have 500 meters worth of "height" removed from it. So force pushing down = 1 bar and (force pushing down) - 500 meters = answer.
So (Difference in force) = (density)(g)(height difference) where given density is in Kg/m^3, height is in m and g is m/s^2. Summing up these values we get Kg/m*s^2 Which is equal to a Newton. The atmospheric pressure is Newtons/m^2.
If I ignore this discrepancy (because the pressure is measuring F/A, and I'm solving just for the force) and solve to problem I end up with a huge number 1.3*8.9*500 which is obviously incorrect. Can someone please steer me in the right direction? thanks

2. Apr 3, 2014

### paisiello2

rho*g*h gives you a pressure not a force.

The question is only asking for the pressure so forget everything about forces. Can you convert 1 bar into kPa?

3. Apr 3, 2014

### Staff: Mentor

To elaborate on what paisiello said, your 1.3*8.9*500 is N/m2, or Pa. This is the difference between the pressure at ground level and the pressure at 500 m. What is 1 bar in Pa?

Chet

4. Apr 3, 2014

### edanzig

1 bar in Pa is approx is 100,000. The difference is 6370. Therefore the answer is .93 bar. (thanks) Just not sure when you say that rho*g*h gives pressure and not force, the units sum to "N" and not "N/m^2."

5. Apr 3, 2014

### paisiello2

If you do the calculation with all of the units then you will see that it gives you a pressure F/A. And you don't sum the units, you multiply them.

6. Apr 3, 2014

### Staff: Mentor

$$\frac{kg}{m^3}\frac{m}{s^2}m=\frac{kgm}{m^2s^2}=\frac{N}{m^2}$$